991微甲01-05班期末考解答和評分標準
1. (10%) Evaluate the intergal
∫
x2ln (x2− 1) dx = .
Sol:
∫
x2ln(x2− 1)dx = x3
3 ln(x2− 1) −
∫ x3 3
2x
x2− 1dx (4 pts)
= x3
3 ln(x2− 1) − 2 3
∫ [
(x2+ 1) + 1 x2− 1
]
dx (6 pts)
= x3
3 ln(x2− 1) − 2
9x3− 2 3x + 1
3ln|x + 1| − 1
3ln|x − 1| + C 2. (10%) Solve the initial value problem y0 = xe− sin x− y cos x, y(0) = 1
2.
Ans: .
Sol:
I(x) = esin x (4 pts)
⇒ yesin x= x2
2 + C (6 pts)
⇒ y = e− sin x(x2 2 +1
2)
3. (10%) Let y = y(x) be a differentiable function on [0,∞) which satisfies y(x) = 1 2−
∫ √x
0
t3y(t2) dt.
(a) Determine a first order differential equation with an initial condition that y(x) satisfies.
Ans: with initial condition .
(b) Solve this differential equation.
Ans: y(x) = .
Sol:
(a) step1: By chain rule and fundamental theorem of calculus I:
y0 =−x32y(x) 2√
x =−1
2xy(x). (3 pts) step2: By plugining x=0 into the integral equation, we have
y(0) = 1
2. (2 pts)
(b) step3: Note that the differential equation in (a) is separable, it follows that
∫ 1 ydy =
∫
−1 2xdx which becomes
ln y =−x2 4 + C.
And the initial condition yields the result C = ln1
2. (5 pts) 4. (10%) Evaluate the improper integral
∫ ∞
1
x
x8+ 4dx = .
Sol:
∫ ∞
1
x
x8+ 4dx = lim
t→∞
∫ t
1
x
x8+ 4dx (1 pt)
(let u = x2) = lim
t→∞
1 2
∫ t2
1
du u4 + 4
= lim
t→∞
1 2
∫ t2
1
du
(u2+ 2u + 2)(u2− 2u + 2) (2 pts)
= lim
t→∞
1 2
∫ t2
1
1
8( u + 2
u2+ 2u + 2− u− 2
u2− 2u + 2)du (2 pts)
= lim
t→∞
1 16
∫ t2
1
(1 2
2u + 2
u2 + 2u + 2+ 1
(u + 1)2+ 1 − 1 2
2u− 2
u2− 2u + 2 + 1
(u− 1)2+ 1)du
y = u2+ 2u + 2 z = u2− 2u + 2
= limt→∞( 1 16
∫ t4+2t2+2
5
1 2
dy y − 1
16
∫ t4−2t2+2
1
1 2
dz z
+ 1
16tan−1(u + 1)¯¯
¯¯t
2
1
+ 1
16tan−1(u + 1)¯¯
¯¯t
2
1
)
= lim
t→∞( 1 32ln y¯¯
¯¯t
4+2t2+2
5
− 1 32ln z¯¯
¯¯t
4−2t2+2
1
) + 1
16(tan−1(t2+ 1) + tan−1(t2− 1) − tan−12− tan−10))
= lim
t→∞( 1
32(lnt4+ 2t2+ 2
t4− 2t2+ 2 − ln 5) + 1
16(tan−1(t2+ 1) + tan−1(t2− 1) − tan−12))
= 1
32(ln 1− ln 5) + 1 16(π
2 + π
2 − tan−12)
= 1
16π− 1
32ln 5− 1
16tan−12 (5 pts)
5. (15%) Let R be the region {
(x, y)|0 ≤ y ≤ 1
xp, x≥ 1} .
(a) Rotate R about the line y =−a, a > 0. For what values of p is the volume of the resulting
solid finite? Ans: . For such p, the volume is .
(b) Rotate R about the line y = 0. For what values of p is the resulting surface area finite?
Ans: .
Sol: (15 % = 6+6+3 ) (a) The volume is
(1)
V =
∫ ∞
1
[ π(1
xp + a)2−πa2]
dx = π
∫ ∞
1
( 1 x2p +2a
xp )
dx
It converges if and only if both integrals converge. Thus, the volume of the resulting solid is finite for
(2) p > 1, ====> 6 points and the volume is
V = π
( 1
2p− 1 + 2a p− 1
)
3 points (b) The surface area is
(3)
2π
∫ ∞
1
1 xp
√
1 + ( −p xp+1)2dx By Comparison Test
1≤√
1 + (...)2 ≤ c
or Limit Comparison Test the above integral converges if and only if
∫ ∞
1
x−pdx converges.
Thus the resulting surface area is finite for (4) p > 1, ====> 6 points
Remark:
1. Please check the calculation that if they derive the expressions (1) and /or (3) first. If there are no (1) and / or (3), even the answers are correct, they get zero points.
2. Give 4 points if the answer of the first p in (a) is incorrect but equation (1) is correct.
3. Give 4 points if the answer of the second p in (b) is incorrect but equation (3) is correct.
6. (15%) (a) Sketch the curve with the polar equation r = 1 + 2 cos 2θ.
(b) Find the intersections of r = 1 and r = 1 + 2 cos 2θ.
Ans: .
(c) Find the area of the region that is inside the curve r = 1 and outside the curve
r = 1 + 2 cos 2θ. Ans: .
Sol:
(a) (4pts)
(b) Solve the equation:
1 + 2 cos 2θ = 1 cos 2θ = 0
θ =±π 4,±3π
4 ,±π
2. (4pts)
(c) By symmetry, the area is:
= 4[π 8 −
∫ π
2
π 4
1
2(1 + 2 cos 2θ)dθ] (3pts)
= 4[π 8 − 1
2
∫ π
2
π 4
1 + 4 cos 2θ + 4 cos22θdθ]
= 4[π 8 − 1
2
∫ π
2
π 4
1 + 4 cos 2θ + 4· 1 + cos 4θ
2 dθ] (2pts)
= 4[π
8 − (3θ + 2 sin 2θ + 1
2sin 4θ)|ππ2
4]
= 4− π. (2pts)
7. (15%) A hypocycloid is a curve traced out by a fixed point P on a circle C of radius b as C
position of P is (a, 0) and the parameter θ is chosen as in the figure (i.e. θ is the angle between the line connecting the centers of the circles and the x-axis.)
(a) Let ϕ be the angle as shown in the figure on the right.
Express ϕ in terms of θ. Ans: .
(b) It is known that the parametric equation of the hypocycloid
is
x(θ) = C1cos θ + C2cos
(a− b b θ
)
y(θ) = C3sin θ + C4sin
(a− b b θ
) ,
where C1, C2, C3, and C4 are constants.
Find C1 = , C2 = , C3 = , C4 = .
(c) For a = 7, b = 3 and 0≤ θ ≤ 2π, the arc length of the hypocycloid is . Sol:
(a) (3pt) ϕ = −a− b
b θ or a− b
b θ The two answers are both ok.
(b) (4pt) C1 = a− b , C2 = b , C3 = a− b , C4 = −b . Every correct coefficient deserves 1pt. Or you may write as the following formula in parameter θ.
x(θ) = (a− b) cos θ + b cos(a− b)θ b y(θ) = (a− b) sin θ − b sin(a− b)θ
b
(c) (8pt) Each step in the following deserves 2pt. If the students have some wrong coefficients in (b), there will still be some process earning part of the points.
For a = 7, b = 3 and 0≤ θ ≤ 2π, the arc length of the hypocycloid is?
Step I Write down the formula and calculate the derivatives of x and y (2pt)
L =
∫ 2π 0
√x0(θ)2 + y0(θ)2 dθ
Step II Simplify the integrant (2pt)
=
∫ 2π
0
√
32− 32 cos7 3θ dθ Step III Use the half-angle formula (2pt)
=
∫ 2π 0
8
√ sin2 7
6θ dθ =
∫ 2π 0
8¯¯
¯¯sin7 6θ¯¯
¯¯ dθ
Step IV Divide [0, 2π] into subintervals to calculate the integral of absolute value of sine (2pt)
= 8
[∫ 6π/7 0
sin7
6θ dθ−
∫ 12π/7 6π/7
sin7
6θ dθ +
∫ 14π/7 12π/7
sin7 6θ dθ
]
= 8·6
7(2+2+1−1/2) = 216 7 Some common mistake and the points:
– something wrong happens in calculation of x0 and y0: 0pt.
– use wrong coefficients in (b) and work out the outcome of Step II: 3pts.
– since the integrant is not in period π/2 or π, so this is wrong:
(X)
∫ 2π
0
√x02+ y02 dθ = 2
∫ π
0
√x02+ y02 dθ
(X)
∫ 2π
0
√x02+ y02 dθ = 4
∫ π/2
0
√x02+ y02 dθ
if the process is correct, it deserves 3pts.
8. (15%) Let C be the curve defined by x(x2+ y2) = (x2− y2).
(a) Find parametric equations of C. (Hint: Let y = tx.)
Ans: x = ,
y = .
(b) There are two tangents at the origin. The equations of these
two tangents are .
(c) The area of the region enclosed by the loop of this curve is . Sol:
(a)
x3(1 + t2) = x2(1− t2) x = 0, (x, y) = (0, 0)
x6= 0, x = 1− t2
1 + t2, y = t1− t2 1 + t2
⇒ x = 1− t2
1 + t2, y = t1− t2
1 + t2, (2 pts each)
(b)
dy dx =
dy dt dx dt
=
−2t4−4t2+2 (1+t2)2
−4t (1+t2)2
= −2t4− 4t2+ 2
−4t , (3 pts) (x, y) = (0, 0)⇔ t = 1ort = −1
t = 1,dy
dx = 1, tanget line : y = x, (1pt) t =−1,dy
dx =−1, tangent line : y = −x, (1pt) (c) if we consider the graph above the x-axis, say y = f (x) > 0 for 0≤ x ≤ 1
∵ x = 1− t2
1 + t2 ⇒ t2(1 + x) = (1− x) ⇒ t =
√1− x 1 + x f (x) = y = tx = x
√1− x 1 + x A = 2
∫ 1 0
y dx = 2
∫ 1 0
f (x) dx = 2
∫ 1 0
x
√1− x
1 + x dx = 2
∫ 1 0
x√ 1− x2 1 + x dx (let x = sin θ, dx = cos θdθ,√
1− x2 = cos θ)
A = 2
∫ π
2
0
sin θ cos θ
1 + sin θ cos θ dθ = 2
∫ π
2
0
sin θ(1− sin2θ)
1 + sin θ dθ = 2
∫ π
2
0
(1− sin θ) sin θ dθ
= 2
∫ π
2
0
(sin θ− sin2θ) dθ = 2
∫ π
2
0
(sin θ−1 2 +1
2cos 2θ) dθ
= 2(− cos θ − 1 2θ + 1
4sin 2θ)¯¯¯
π 2
0
= 2(−π
4 − (−1)) = 2 − π 2
You should get 1 points if you have written down the range of integration correctly.
2 points more if the integrand is valid.
Finally, 3 points for the answer.
